16 DISCRETE-SIGNAL ANALYSIS AND DESIGN part and an imaginary part, the real parts add coherently and the imaginary parts add coherently, and the power is complex (real watts and imaginary vars). There is much more about this later. If K x =1.2 in Eq. (1-1), then 1.2 cycles would be visible, the spectrum would contain many frequencies, and the Þnal phase would change to (0.2 · 2π) radians. The value of the phase angle in degrees for each complex X (k)is φ(k) = arctan Im X(k) Re X(k) · 180 π degrees (1-3) For an example of this type of sequence, look ahead to Fig. 1-6. A later section of this chapter gives more details on complex frequency-domain sequences. At this point, notice that the complex term exp(j ωt) is calculated by Mathcad using its powerful and efÞcient algorithms, eliminating the need for an elaborate complex Taylor series expansion by the user at each value of (n)or(ω). This is good common sense and does not derail us from our discrete time/frequency objectives. At each (k) stop, the sum is performed at 0 to N −1 values of time (n), for a total of N values. It may be possible to evaluate accurately enough the sum at each (k) value with a smaller number of time steps, say N /2 or N /4. For simplicity and best accuracy, N will be used for both (k) and (n). Using Mathcad to Þnd the spectrum without assigning discrete (k) values from 0 to N −1, a very large number of frequency values are evaluated and a continuous graph plot is created. We will do this from time to time, and the summation () becomes more like an integral , but this is not always a good idea, for reasons to be seen later. Note also that in Eq. (1-2) the factor 1/N ahead of the sum and the minus sign in the exponent are used but are not used in Eq. (1-8) (look ahead). This notation is common in engineering applications as described by [Ronald Bracewell, 1986] and is also an option in Mathcad (functions FFT and IFFT). See also [Oppenheim and Willsky et al., 1983, p. 321]. This agrees with the practical engineering emphasis of this book. It also agrees with our assumption that each record, 0 to N −1, is one replication of an inÞnite steady-state signal. These two equations, used together and consistently, produce correct results. FIRST PRINCIPLES 17 Each (k ) is a harmonic number for the frequency sequence X (k). To repeat a few previous statements for emphasis, k =1 is the fundamen- tal frequency, k =2 is second harmonic, etc. A two-sided (positive and negative) phasor spectrum is produced by this equation (we will learn to appreciate the two-sided spectrum concept). N , an integral power of 2, is chosen large enough to provide adequate resolution of the spectrum (sufÞcient harmonics of k =1). The dc component is at k =0[wherethe exp(0) term =1.0] and X(0) = 1 N N−1 n=0 x(n) =x(n) volts (1-4) which is the time average over the entire sequence, 1.0, in Fig. 1-2. Equation (1-2) can be used directly to get the spectrum, but as a matter of considerable interest later it can be separated into two regions having an equal number of data points, from 0 to N /2 −1 and from N /2 to N −1 as shown in Eq. (1-5). If N =8, then k (positive frequencies) =1, 2, 3 and k (negative frequencies) =7, 6, 5. Point N is the beginning of the next periodic continuation. Dc is at k =0, and N /2 is not used, for reasons to be explained later in this chapter. Consider the following manipulations of Eq. (1-2): X(k) = 1 N ⎡ ⎣ N/2−1 n=0 x(n)e −jk2π ( n N ) + N−1 n=N/2 x(n)e −jk2π ( n N ) ⎤ ⎦ (1-5) The last exponential can be modiÞed as follows without changing its value: e −jk2π n N = e j(2πn) 360 ◦ e −jk2π n N = e j2πn 1− k N = e j2π ( N−k ) n N (1-6) and Eq. (1-2) becomes X(k) = 1 N ⎡ ⎣ N/2−1 n=0 x(n)e −jk2π n N + N−1 n=N/2 x(n)e j2π(N−k) n N ⎤ ⎦ (1-7) 18 DISCRETE-SIGNAL ANALYSIS AND DESIGN The second exponential is the phase conjugate (e −jθ →e +jθ )oftheÞrst and is positioned as shown in Fig. 1-2b for k =N /2 to N −1. At k =0 we see the dc. The two imaginary components −j 0.5 and +j 0.5, are at k =1andk =63 (same as k =−1), typical for a sine wave of length 64. We use this method quite often to convert two-sided sequences into one-sided (positive-time or positive-frequency) sequences (see Chapter 2 for more details). INVERSE DISCRETE FOURIER TRANSFORM The inverse transformation (IDFT) in Eq. (1-8) [Oppenheim et al., 1983, p. 321] takes the two-sided spectrum X (k ) in Fig. 1-2b and exactly recre- ates the original two-sided time sequence x(n) shown in Fig. 1-2c: x(n) = N−1 k=0 X(k)e jk2π ( n N ) (1-8) At each value of (n) the spectrum values X (k ) are summed from k =0to k =N −1. In Eq. (1-8) the phase increments are in the counter-clockwise (positive) direction. This reverses the negative phase increments that were introduced into the DFT [Eq. (1-2)]. This step helps to return each complex X (k ) in the frequency domain to a real x(n) in the time domain. See further discussion later in the chapter. It is interesting to focus our attention on Eqs. (1-2) and (1-8) and to observe that in both cases we are simultaneously in the time and frequency domains. We must have data from both domains to travel back and forth. This conÞrms that we are learning to be comfortable in both domains at once, which is exactly what we need to do. So far, Eqs. (1-2) and (1-8) have been used directly, without any need for a faster method, the FFT (the Fast Fourier Transform), described later. Modern personal computers are usually fast enough for simple problems using just these two equations. Also, Eqs. (1-2) and (1-8) are quite accurate and very easy to use in computerized analysis (however, Mathcad also has very excellent tools for numerical and symbolic integration that we will use frequently). We do not have to worry about those two discrete FIRST PRINCIPLES 19 equations in our applications because they have been thoroughly tested. It is a good idea to use Eqs. (1-2) and (1-8) together as a pair. To narrow the time or frequency resolution, multiply the value of N by 2 M (m =1, 2, 3, ), as shown in the next section. FREQUENCY AND TIME SCALING Suppose a signal spectrum extends from 0 Hz to 30 MHz (Fig. 1-3) and we want to display it as a 32-point (=2 5 ) two-sided spectrum. The positive side of the spectrum has 15 X (k ) values from 1 to N /2 −1 (not count- ing 0 and N /2), and the negative side of the spectrum also has 15 X (k ) values from N /2 +1toN −1 (not counting N /2 and N ). The frequency range 0 to 30 MHz consists of a fundamental frequency k 1 and 2 4 −1 =15 harmonics of k 1 . The fundamental frequency k1 is determined by k 1 ·15 = 3·10 7 ∴ k 1 = 3·10 7 15 = 2 MHz (1-9) and this is the best resolution of frequency that can be achieved with 15 points (positive or negative frequencies) of a 30-MHz signal using a 32-point two-sided spectrum. If we use 2048 data points, we can get 29.31551-kHz resolution using Eq. (1-9). +/− 30 MHz 0 0 2.0 MHz resolution K= +1 K = −1 −2 MHz Figure 1-3 A 30-MHz two-sided spectrum with 32 frequency samples, including 0. 20 DISCRETE-SIGNAL ANALYSIS AND DESIGN An excellent way to improve this example is to frequency-convert the signal band to a much lower frequency, for example 3 MHz, using a very stable local oscillator, which would give us a 2931.55- Hz resolution for this example. Increasing the samples to 2 14 at 3 MHz provides a resolution of 366.26 Hz, and so forth for higher sample numbers. This is basically what spectrum analyzers do. The good news for this problem is that a hardware frequency translator may not be necessary. If the signal is narrowband, such as speech or low-speed data or some other bandlimited process, the original 30-MHz problem might be restated at 3 MHz, or maybe even at 0.3 MHz, with the same signal bandwidth and with no loss of correct results, but with greatly improved resolution. With programs for personal computer analysis, very large numbers of samples are not desirable; therefore, we do not try to push the limits too much. The waveform analysis routines usually tell us what we want to know, using more reasonable numbers of samples. Designing the frequency and time scales is very helpful. Consider a time scaling example, a sequence (record length) that is 10 μsec long from start of one sequence to the start of the next sequence, as shown in Fig. 1-4. For N =4 there are 4 time values (0, 1, 2, 3) and 4 time intervals (1, 2, 3, 4) to the beginning of the next sequence, which is 10 −5 /4 =2.5 μsec per interval. In the Þrst half there are 2 intervals for a total of 5.0 μsec. For the second half there are also 2 intervals, for a total of 5.0 μsec. Each interval is a “band” of possibly smaller time increments. The total time is 10.0 μsec. 01 2 3 1 2 3 4 0 2.5 +5.0 − 5.0 −2.5 0 N = 4 msec 10 msec Figure 1-4 A 10-μsec time sequence with positive and negative time values. . 16 DISCRETE-SIGNAL ANALYSIS AND DESIGN part and an imaginary part, the real parts add coherently and the imaginary parts add coherently, and the power is complex (real watts and imaginary vars) e j2πn 1− k N = e j2π ( N−k ) n N (1-6) and Eq. (1-2) becomes X(k) = 1 N ⎡ ⎣ N/2−1 n=0 x(n)e −jk2π n N + N−1 n=N/2 x(n)e j2π(N−k) n N ⎤ ⎦ (1-7) 18 DISCRETE-SIGNAL ANALYSIS AND DESIGN The second exponential. attention on Eqs. (1-2) and (1-8) and to observe that in both cases we are simultaneously in the time and frequency domains. We must have data from both domains to travel back and forth. This conÞrms